Question
CSATMediumPrelims 2022Maths

Which number amongst 2402^{40}, 3213^{21}, 4184^{18} and 8128^{12} is the smallest?

Explanation

The given numbers 2402^{40}, 3213^{21}, 4184^{18}, and 8128^{12} can be simplified as follows:

2402^{40} remains as 2402^{40} 3213^{21} remain as 3213^{21} 418=2364^{18} = 2^{36} 812=2368^{12} = 2^{36}

This simplifies our choices to 2402^{40} , 3213^{21}, 2362^{36} , and 2362^{36}. Since 2402^{40} is obviously larger than 2362^{36} , it cannot be the smallest. Additionally, because we cannot have two correct answers, we are left to consider 3213^{21} as the smallest option.

If further comparison were needed between 2362^{36} and 3213^{21}, we could factorize them for an estimate without full calculation:

2362^{36} can be written as 2×235=2×(25)7=2×3272 \times 2^{35} = 2 \times (2^{5})^{7} = 2 \times 32^{7} 321=(33)7=2773^{21} = (3^{3})^{7} = 27^{7}

From observation, 3213^{21} is smaller than 2362^{36} without detailed computation, as the exponential growth in the latter is more significant.

Maths PYQs from 2022

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